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What is the change in enthalpy (ΔH) for the combustion of methane (CH4) at standard conditions?

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The combustion of methane is an exothermic reaction, which means it releases heat. The balanced chemical equation for the combustion of methane is:CH g  + 2O g   CO g  + 2HO l To calculate the change in enthalpy  H  for this reaction, we need to consider the standard enthalpies of formation  Hf  for each of the compounds involved. The standard enthalpy of formation is the change in enthalpy when one mole of a substance is formed from its elements in their standard states.H = [Hf CO  + 2Hf HO ] - [Hf CH  + 2Hf O ]The standard enthalpies of formation for the compounds are:Hf CH  = -74.8 kJ/molHf O  = 0 kJ/mol  since O is in its standard state Hf CO  = -393.5 kJ/molHf HO  = -285.8 kJ/molNow, we can plug these values into the equation:H = [ -393.5  + 2 -285.8 ] - [ -74.8  + 2 0 ]H =  -393.5 - 571.6  -  -74.8 H = -965.1 + 74.8H = -890.3 kJ/molSo, the change in enthalpy  H  for the combustion of methane  CH  at standard conditions is approximately -890.3 kJ/mol.

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